Question

10. A block of mass 1 kg is at rest on a horizontal table. The
coefficient of static friction between the block and the
table is 0.5. If g = 10 ms 2, then the magnitude of the force
acting upwards at an angle of 60° from the horizontal that
will just start the block moving is
(a) 5 N
(b) 5.36 N
(c) 74.6 N
(d) 10 N
tehody slide down a 45° inclined nlane​

Answer 1

given m=1kg, weight = mg =10N

R=normal reaction of table on block

coefficient of static friction be u=0.5

applied force(F) splits into 2components..

Fcos60(Horizontal component)

Fsin60(Vertical component along direction of weight)

R=mg+Fsin60

=10+Fsin60

now equating all forces..

Fcos60=uR

Fcos60=(0.5)[mg+Fsin60]

Fcos60=(0.5)[10+Fsin60]

solving F=74.6N

Answer 2

Explanation:

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